3.11.35 \(\int \frac {x^2}{(c+a^2 c x^2)^{5/2} \arctan (a x)^{3/2}} \, dx\) [1035]

3.11.35.1 Optimal result

Integrand size = 26, antiderivative size = 281 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=-\frac {2 x^2}{a c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}}-\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^3 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {2 \pi }{3}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{a^3 c^2 \sqrt {c+a^2 c x^2}} \]

output

1/2*FresnelS(6^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2 
+1)^(1/2)/a^3/c^2/(a^2*c*x^2+c)^(1/2)-1/2*FresnelS(2^(1/2)/Pi^(1/2)*arctan 
(a*x)^(1/2))*2^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^3/c^2/(a^2*c*x^2+c)^(1/2 
)-2*x^2/a/c/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(1/2)
 

3.11.35.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\frac {-\frac {12 a^2 x^2}{\sqrt {\arctan (a x)}}+\sqrt {6 \pi } \left (1+a^2 x^2\right )^{3/2} \left (-3 \sqrt {3} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )-\frac {\left (1+a^2 x^2\right )^{3/2} \left (3 \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+3 \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )+\sqrt {3} \left (\sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+\sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )\right )\right )}{\sqrt {\arctan (a x)}}}{6 a^3 c \left (c+a^2 c x^2\right )^{3/2}} \]

input

Integrate[x^2/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]^(3/2)),x]
 

output

((-12*a^2*x^2)/Sqrt[ArcTan[a*x]] + Sqrt[6*Pi]*(1 + a^2*x^2)^(3/2)*(-3*Sqrt 
[3]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] + FresnelS[Sqrt[6/Pi]*Sqrt[ArcT 
an[a*x]]]) - ((1 + a^2*x^2)^(3/2)*(3*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I 
)*ArcTan[a*x]] + 3*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]] + Sqrt[3] 
*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] + Sqrt[I*ArcTan[a* 
x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])))/Sqrt[ArcTan[a*x]])/(6*a^3*c*(c + a^2* 
c*x^2)^(3/2))
 

3.11.35.3 Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.83, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5503, 5506, 5505, 3042, 3793, 2009, 4906, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2}{\arctan (a x)^{3/2} \left (a^2 c x^2+c\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 5503

\(\displaystyle \frac {4 \int \frac {x}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a}-2 a \int \frac {x^3}{\left (a^2 c x^2+c\right )^{5/2} \sqrt {\arctan (a x)}}dx-\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 5506

\(\displaystyle \frac {4 \sqrt {a^2 x^2+1} \int \frac {x}{\left (a^2 x^2+1\right )^{5/2} \sqrt {\arctan (a x)}}dx}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {2 a \sqrt {a^2 x^2+1} \int \frac {x^3}{\left (a^2 x^2+1\right )^{5/2} \sqrt {\arctan (a x)}}dx}{c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 5505

\(\displaystyle \frac {4 \sqrt {a^2 x^2+1} \int \frac {a x}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \int \frac {a^3 x^3}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 \sqrt {a^2 x^2+1} \int \frac {a x}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \int \frac {\sin (\arctan (a x))^3}{\sqrt {\arctan (a x)}}d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {4 \sqrt {a^2 x^2+1} \int \frac {a x}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \int \left (\frac {3 a x}{4 \sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}-\frac {\sin (3 \arctan (a x))}{4 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 \sqrt {a^2 x^2+1} \int \frac {a x}{\left (a^2 x^2+1\right )^{3/2} \sqrt {\arctan (a x)}}d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^3 c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {4 \sqrt {a^2 x^2+1} \int \left (\frac {a x}{4 \sqrt {a^2 x^2+1} \sqrt {\arctan (a x)}}+\frac {\sin (3 \arctan (a x))}{4 \sqrt {\arctan (a x)}}\right )d\arctan (a x)}{a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^3 c^2 \sqrt {a^2 c x^2+c}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 x^2}{a c \sqrt {\arctan (a x)} \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 \sqrt {a^2 x^2+1} \left (\frac {3}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^3 c^2 \sqrt {a^2 c x^2+c}}+\frac {4 \sqrt {a^2 x^2+1} \left (\frac {1}{2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )+\frac {1}{2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )\right )}{a^3 c^2 \sqrt {a^2 c x^2+c}}\)

input

Int[x^2/((c + a^2*c*x^2)^(5/2)*ArcTan[a*x]^(3/2)),x]
 

output

(-2*x^2)/(a*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]]) - (2*Sqrt[1 + a^2*x 
^2]*((3*Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/2 - (Sqrt[Pi/6] 
*FresnelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/2))/(a^3*c^2*Sqrt[c + a^2*c*x^2]) 
 + (4*Sqrt[1 + a^2*x^2]*((Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]] 
])/2 + (Sqrt[Pi/6]*FresnelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/2))/(a^3*c^2*Sq 
rt[c + a^2*c*x^2])
 

3.11.35.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 5503

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(q_), x_Symbol] :> Simp[x^m*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p 
+ 1)/(b*c*d*(p + 1))), x] + (-Simp[c*((m + 2*q + 2)/(b*(p + 1)))   Int[x^(m 
 + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] - Simp[m/(b*c*(p + 
1))   Int[x^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x]) /; F 
reeQ[{a, b, c, d, e, m}, x] && EqQ[e, c^2*d] && IntegerQ[m] && LtQ[q, -1] & 
& LtQ[p, -1] && NeQ[m + 2*q + 2, 0]
 

rule 5505

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(q_), x_Symbol] :> Simp[d^q/c^(m + 1)   Subst[Int[(a + b*x)^p*(Sin[x]^m/ 
Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p 
}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q 
] || GtQ[d, 0])
 

rule 5506

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(q_), x_Symbol] :> Simp[d^(q + 1/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]) 
  Int[x^m*(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, 
d, e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(I 
ntegerQ[q] || GtQ[d, 0])
 

3.11.35.4 Maple [F]

\[\int \frac {x^{2}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \arctan \left (a x \right )^{\frac {3}{2}}}d x\]

input

int(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(3/2),x)
 

output

int(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(3/2),x)
 

3.11.35.5 Fricas [F(-2)]

Exception generated. \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\text {Exception raised: TypeError} \]

input

integrate(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(3/2),x, algorithm="fricas")
 

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

3.11.35.6 Sympy [F]

\[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\int \frac {x^{2}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}\, dx \]

input

integrate(x**2/(a**2*c*x**2+c)**(5/2)/atan(a*x)**(3/2),x)
 

output

Integral(x**2/((c*(a**2*x**2 + 1))**(5/2)*atan(a*x)**(3/2)), x)
 

3.11.35.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

input

integrate(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(3/2),x, algorithm="maxima")
 

output

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

3.11.35.8 Giac [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\text {Timed out} \]

input

integrate(x^2/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(3/2),x, algorithm="giac")
 

output

Timed out
 

3.11.35.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \arctan (a x)^{3/2}} \, dx=\int \frac {x^2}{{\mathrm {atan}\left (a\,x\right )}^{3/2}\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

input

int(x^2/(atan(a*x)^(3/2)*(c + a^2*c*x^2)^(5/2)),x)
 

output

int(x^2/(atan(a*x)^(3/2)*(c + a^2*c*x^2)^(5/2)), x)